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For example, it takes a car four times the kinetic energy to travel the same distance at twice the speed, which quadruples the drag experienced by the car. This observation is described by the standard equations:

Drag = 0.5 (Velocity2 x Air Density x Surface Area x Coefficient of Drag)

K.E. = 0.5mv2

I'd really appreciate it if someone could critique a new explanation based on Newtons second law (force = ma), or confirm if this explanation is correct:

For a car travelling at a constant velocity. The force from the engine pushing the car forwards (i.e. Force FORWARD = ma = m/dt x dv), equals mass of air each second (m/dt) that the car travels through and accelerates away from it at a given velocity (dv). The more aerodynamic cars will have a lower 'dv'.

Force FORWARD = ma = m x dv/dt = m/dt x dv

Just to be clear, note that 'm' and 'v' are NOT not the mass or the velocity of the car.

This action generates an equal and opposite force, called drag. At constant speed, so no acceleration, the forward force will equal the drag, which is the force required to push the air out of the car's path.

Force FORWARD = Force DRAG

enter image description here

If the car's speed doubles (x2); then the forward force, drag and kinetic energy required all quadruple (x4), as described by their standard equations above. The explanation for this quadrupling is:

(i) The car will travel through twice the mass of air each second (2x ‘m/dt’).

(ii) As the car's momentum has doubled, the car will accelerate this air travelled through to twice the velocity as before (2x ‘dv’).

The combined effect is to quadruple the total force for drag each second; (i.e. 4x Force DRAG = 2m/dt x 2dv).

enter image description here

In addition, force and energy are proportional for a given distance travelled by the car, based on their respective definitions:
Force = 1 kg m /s2 = 1 N
Energy = 1 kg m2 /s2 = 1 J

Given that the distance travelled by the car has not changed, but the force required to push the car forwards and drag have both quadrupled (x4), therefore, the kinetic energy required will also quadruple (KE x 4).

This is why it takes four times the kinetic energy and drag to travel the same distance at twice the speed; And why kinetic energy and drag are proportional to velocity2.

Is this correct? Thanks!

G. Smith
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Nick Landell
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    The questions about kinetic energy and drag are two separate questions. For the question of why kinetic energy varies as $v^2$, that is probably a duplicate of this question. Also, drag is not always proportional to $v^2$, as explained here. – Sandejo Oct 15 '20 at 05:07
  • Hi Nick Landell. Welcome to Phys.SE. To reopen this post (v2), consider for starters to remove the duplicate question. – Qmechanic Oct 15 '20 at 09:49
  • I posted them here because drag and kinetic energy are related. Both are proportional to velocity 2. They are not separate. Thanks. – Nick Landell Oct 16 '20 at 08:26

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